JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A square loop of side \(2.0\,cm\) is placed inside a long solenoid that has \(50\) turns per centimetre and carries a sinusoidally varying current of amplitude \(2.5\,A\) and angular frequency \(700\,rad\,s ^{-1}\). The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is \(x \times 10^{-4} V\). The value of \(x\) is \(.........\)(Take, \(\pi=\frac{22}{7}\))
- A \(43\)
- B \(42\)
- C \(44\)
- D \(41\)
Answer & Solution
Correct Answer
(C) \(44\)
Step-by-step Solution
Detailed explanation
\(B _{\text {due to solenoid }}=\mu_0 nI\) \(\Phi_{\text {through square }}=\mu_0 nI \times A \quad( A =\text { Area })\) \(Emf =\mu_0 nA \times \frac{ dI }{ dt }\) \(=\mu_0 n A \times I _0 \omega \cos \omega t\) \(\text { Emf amplitude }=\mu_0 n A \times I _0 \omega\)…
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